write the empirical formula for at least four ionic compounds that could be formed from the following ions: pb^{4+}, no_{3}^{-}, io_{3}^{-}, fe^{3+}

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Accepted Answer

# Explanation:
## Step1: Combine $Pb^{4 +}$ with $NO_{3}^{-}$
The charge of $Pb^{4+}$ is $+ 4$ and the charge of $NO_{3}^{-}$ is $-1$. To balance the charges, we need $4$ $NO_{3}^{-}$ ions for each $Pb^{4+}$ ion. So the formula is $Pb(NO_{3})_{4}$.
## Step2: Combine $Pb^{4 +}$ with $IO_{3}^{-}$
The charge of $Pb^{4+}$ is $+4$ and the charge of $IO_{3}^{-}$ is $-1$. To balance the charges, we need $4$ $IO_{3}^{-}$ ions for each $Pb^{4+}$ ion. So the formula is $Pb(IO_{3})_{4}$.
## Step3: Combine $Fe^{3 +}$ with $NO_{3}^{-}$
The charge of $Fe^{3+}$ is $+3$ and the charge of $NO_{3}^{-}$ is $-1$. To balance the charges, we need $3$ $NO_{3}^{-}$ ions for each $Fe^{3+}$ ion. So the formula is $Fe(NO_{3})_{3}$.
## Step4: Combine $Fe^{3 +}$ with $IO_{3}^{-}$
The charge of $Fe^{3+}$ is $+3$ and the charge of $IO_{3}^{-}$ is $-1$. To balance the charges, we need $3$ $IO_{3}^{-}$ ions for each $Fe^{3+}$ ion. So the formula is $Fe(IO_{3})_{3}$.

# Answer:
$Pb(NO_{3})_{4}$, $Pb(IO_{3})_{4}$, $Fe(NO_{3})_{3}$, $Fe(IO_{3})_{3}$

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Question

Explanation:

Step1: Combine \(Pb^{4 +}\) with \(NO_{3}^{-}\)

Step2: Combine \(Pb^{4 +}\) with \(IO_{3}^{-}\)

Step3: Combine \(Fe^{3 +}\) with \(NO_{3}^{-}\)

Step4: Combine \(Fe^{3 +}\) with \(IO_{3}^{-}\)

Answer: